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I think I understand, but where do you find force?

  1. Nov 30, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the coefficient of kinetic friction that is used to stop a train engine along its track of the trains velocity is 30 m/s when the brakes are applied, and it takes 30 sec for the 10,000 kg train to stop
    2. Relevant equations
    Ffric=MN=Mmg
    (I think that's the equation to use.

    3. The attempt at a solution
    Ffric=M(10000kg)(9.8m/s)
    Ffric=M(98,000)
    What do you use for the force? 30m/s?
     

    Attached Files:

  2. jcsd
  3. Nov 30, 2015 #2

    SammyS

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    Hello Laura0901. Welcome to PF !

    What is the acceleration of the train ?
     
  4. Nov 30, 2015 #3
    It is not said in the problem.
     
  5. Nov 30, 2015 #4

    SammyS

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    Of course you must figure it out from the given information.
     
  6. Nov 30, 2015 #5
    And how would you do that? Using the kinematic equations?
     
  7. Nov 30, 2015 #6
    Why would we need to find the acceleration? I need the friction force.
     
  8. Nov 30, 2015 #7

    SammyS

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    And, just what force is it that is slowing the train?
     
  9. Nov 30, 2015 #8
    Either gravity or velocity..I think.
     
  10. Nov 30, 2015 #9

    SammyS

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    The train slows to a stop. That's an acceleration. That requires a force. (Ask Mr. Newton.)

    Gravity is a force, but it acts in a vertical, the train's motion and its acceleration are in the horizontal direction. Gravity is not the force which slows the train.

    Velocity is not a force.
     
  11. Nov 30, 2015 #10
    1) Figure out the force in the x-direction that SammyS is talking about, the force that causes the train to stop.
    2) Use kinematics to find a
    3)Use Fx=max to find the magnitude of the force mentioned in step one
    Then you should have enough data to solve Ffric=M(98,000)
     
  12. Nov 30, 2015 #11
    Thanks for trying to help me understand
     
  13. Nov 30, 2015 #12
    There aren't any forces in the x direction. I don't think length is a force. I don't know how to find friction.
     
  14. Nov 30, 2015 #13

    SammyS

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    Then the train's going to keep going in the x direction.

    By the way: What direction is the friction force ?
     
  15. Nov 30, 2015 #14
    The Positive direction
     
  16. Nov 30, 2015 #15
    Is it 2N?
     
  17. Nov 30, 2015 #16

    SammyS

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    Positive in what sense?
    Positive x-direction?
    or
    Positive as in upward?
     
  18. Nov 30, 2015 #17
    The x direction, it is horizontal
     
  19. Nov 30, 2015 #18
    Okay... I hope you are asking if fk is 2N because M is not in newtons. I don't think fk=2N either though...
     
  20. Nov 30, 2015 #19

    SammyS

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    Right. It would take very long time to stop.
     
  21. Nov 30, 2015 #20
    Ughh I'm so bad at everything sorry guys.
     
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